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I need a hint to solve this:

Let $a_1 = 1$ and define a sequence recursively by $$a_{n+1} = \sqrt{a_1 + a_2 + ... a_n}$$

Show that $$\lim_{n \to \infty} \dfrac{a_n}{n} = \dfrac{1}{2}$$

Any help? Thank you

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marked as duplicate by Sil, Xander Henderson, max_zorn, Namaste, Adrian Keister Aug 12 '18 at 1:06

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    $\begingroup$ Another way of writing the recursion is as follows:$$a_{n+1}=\sqrt{a_n^2+a_n}$$ $\endgroup$ – Akiva Weinberger Oct 2 '14 at 3:09
  • $\begingroup$ Actually that was my 1st try, but I couldn't finish it. Then I used that to get $a_{n+1} \le \sqrt{2} a_n $ but nothing. $\endgroup$ – FormerMath Oct 2 '14 at 3:12
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    $\begingroup$ Use columbus8myhw's hint to prove that $a_n$ increases without bound. Then use the same relation to prove that $a_{n+1} = a_n + 1/2 + O(1/a_n)$. $\endgroup$ – user180040 Oct 2 '14 at 3:17
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    $\begingroup$ Idea: If $a_n\le\frac n2$, then $a_{n+1}=\sqrt{a_n^2+a_n}\le\sqrt{\frac{n^2}4+\frac n2}=\frac{\sqrt{n^2+2n}}2<\frac{\sqrt{n^2+2n+1}}2=\frac{n+1}2$. Since $a_2=1<\frac22$, we know that (for $n\ge2$) $a_n\le\frac n2$. $\endgroup$ – Akiva Weinberger Oct 2 '14 at 3:24
  • $\begingroup$ I will give it a try. Thanks for the help! $\endgroup$ – FormerMath Oct 2 '14 at 3:38
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First, you need to prove that $a_n\to\infty$ as $n\to\infty$. Then, by applying Stolz–Cesàro theorem we have $$\lim_{n\to\infty}\frac{a_n}{n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)-n}=\lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}(\sqrt{a_{n}^2+a_n}-a_n)=\lim_{n\to\infty}\frac{a_n}{\sqrt{a_{n}^2+a_n}+a_n}=\lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{a_n}}+1}=\frac12.$$

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  • $\begingroup$ The first equality is verified only if the limit of the second term exists, so maybe the end should be re-written differently ? Anyway, nice answer, +1. $\endgroup$ – Traklon Oct 2 '14 at 7:17

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